Find the inverse of the matrix, $\text B = \left[\begin{array}{rr}-9 & -6 \\ 0 & 2 \end{array}\right]$. Non-integers should be given either as decimals or as simplified fractions. $ B^{-1}=$
Answer: The Strategy To find the inverse of an invertible matrix, we can use Gaussian Elimination. To do this, we do the following. First, we append the matrix $\text B$ with the identity matrix $\text I$ to get [ B | I ] \left[\begin{array} ~\text B ~ |~\text I\end{array}\right]. Next, we use Gaussian Elimination to reduce $\text B$ to the identity matrix, $\text I$. Performing the same operations on $\text I$ will convert it to $\text B^{-1}$, so that our new matrix becomes [ I | B − 1 ] \left[\begin{array} ~\text I ~ |~\text B^{-1}\end{array}\right]. Appending $\text B$ with $\text I$ [ B | I ] = [ − 9 0 − 6 2 1 0 0 1 ] \left[\begin{array} ~\text B ~ |~\text I\end{array}\right]=\left[\begin{array}{rr}-9 & -6 & 1 & 0 \\ 0 & 2 & 0 & 1 \end{array}\right] Eliminating the leading term in the second row We want the first term of $R_2$ to equal $0$, but since it already is, we don't need to subtract a multiple of $R_1$ from $R_2$ in our first step. Reducing the leading terms and back-solving Now, let's reduce the leading term of $R_2$ to equal $1$. $\left[\begin{array}{rr}-9 & -6 & 1 & 0 \\ {0} & {2} & {0} & {1} \end{array}\right]\xrightarrow{\dfrac{1}{2}R_2\rightarrow R_2}\left[\begin{array}{rr}-9 & -6 & 1 & 0 \\ {0} & {1} & {0} & {\dfrac{1}{2}} \end{array}\right]$ We are ready to back-solve to get [ I | B − 1 ] \left[\begin{array} ~\text I ~ |~\text B^{-1}\end{array}\right]. $\begin{aligned}\!\!\left[\begin{array}{rr}\!{-9}\! & {-6}\! & {1}\! & {0} \\ \!0\! & 1\! & 0\! &\dfrac{1}{2} \!\end{array}\right]\!\xrightarrow{\!\!R_1+6R_2\rightarrow R_1\!\!} \!\!&\left[\begin{array}{rr}\!{\!-9}\!\! & {0}\!\! & {1} \!\!& {3} \!\\ \!0\! & 1\! & 0\! &\dfrac{1}{2} \end{array}\right] \!\!\xrightarrow{\!\!{-}\dfrac{1}{9}R_1\!\rightarrow R_1\!\!}\!\!\left[\begin{array}{rr}{1}\!\! & {0}\!\! & {-\!\dfrac{1}{9}}\!\! & {-\!\dfrac{1}{3}} \\ \!0\! & 1\! & 0\! &\dfrac{1}{2} \end{array}\right]\end{aligned}$ Therefore $\text B^{-1}=\left[\begin{array}{rr} -\dfrac{1}{9}\!\! & -\dfrac{1}{3} \\ 0 \!\!& \dfrac{1}{2} \end{array}\right]$. Summary $\text B^{-1}=\left[\begin{array}{rr} -\dfrac{1}{9}\!\! & -\dfrac{1}{3} \\ 0 \!\!& \dfrac{1}{2} \end{array}\right]$